What is the solution for y^{''}+4y^'=6e^{3t}-3e^{-t} ?

The solution for y^{''}+4y^'=6e^{3t}-3e^{-t} is y=c_{1}+c_{2}e^{-4t}+2/7 e^{3t}+e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'=6e^{3t}-3e^{-t}

y^{^{''}} + 4 y^{^{'}} = 6 e^{3 t} - 3 e^{- t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

Solution steps

y^{''} (t) + 4 y^{'} (t) = 6 e^{3 t} - 3 e^{- t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{2}{7} e^{3 t} + e^{- t}

y^{^{''}} + 4 y = 2 sin (t)

y^{^{''}} - 3 y^{^{'}} - 10 y = 54 e^{7 t}, y (0) = 5

y^{^{''}} + 9 y = - 2 e^{3} t

y^{^{'''}} - 2 y^{^{''}} = 4

y^{^{''}} - 9 y = e^{(- 2 x)} (- 17 + 10 x)

What is the solution for y^{''}+4y^'=6e^{3t}-3e^{-t} ?
The solution for y^{''}+4y^'=6e^{3t}-3e^{-t} is y=c_{1}+c_{2}e^{-4t}+2/7 e^{3t}+e^{-t}